Can you please check my work? I got -4.3KJ/mol
If this is not right can you tell me what I am doing wrong? Thank you!
Questions. 1.a. Use Hess's Law and the measured mean enthalpy changes for the NaOH-HCl and NH3-HCl reactions to calculate the enthalpy change to be expected for the reaction:
NaOH + NH4Cl => NaCl + NH3 + H2O
It's wanting the delta H value.
These are the values from the mean enthalpy.
Delta H for NaOH & HCl -59 KJ/mol
Delta H for NH3 & HCl -50. KJ/mol
Delta H for NaOH & NH4Cl -5.7 KJ/mol
I'm confused by your data.
You've listed dH for the reaction you want as -5.7 kJ/mol. So
a. why do you have the question if the answer is -5.7?
b. Why do you have -4.3 if the data says it is -5.7
If I did this and used ONLY the first two equations you have, I get this.
NaOH + HCl ==> NaCl + H2O dH = -59 kJ/mol
Reverse equation 2 to obtain
NH4Cl ==> HCl + NH3 dH = +50
Add the two equations
NaOH + NH4Cl ==> NaCl + H2O + NH3 and add dH -59 kJ/mol + 50 kJ/mol = -9 kJ/mol
To calculate the enthalpy change for the reaction NaOH + NH4Cl -> NaCl + NH3 + H2O using Hess's Law, we can apply the following steps:
Step 1: Write down the balanced chemical equations for the given reactions and the desired reaction:
1. NaOH + HCl -> NaCl + H2O
2. NH3 + HCl -> NH4Cl
Desired reaction:
3. NaOH + NH4Cl -> NaCl + NH3 + H2O
Step 2: Notice that we need to multiply equation 2 by -1 because the desired reaction involves NH4Cl as a reactant, but the given reaction gives NH4Cl as a product. This allows us to cancel out NH4Cl.
Balanced equations:
1. NaOH + HCl -> NaCl + H2O
2. -NH3 - HCl -> -NH4Cl
Desired reaction:
3. NaOH + NH4Cl -> NaCl + NH3 + H2O
Step 3: Determine the change in enthalpy for each given reaction using the provided values:
ΔH1 = -59 kJ/mol (NaOH + HCl -> NaCl + H2O)
ΔH2 = -50 kJ/mol (NH3 + HCl -> NH4Cl)
Step 4: Flip equation 2, so the NH4Cl is a reactant:
ΔH2' = 50 kJ/mol (NH4Cl -> NH3 + HCl)
Step 5: Multiply equation 2 by the factor required to balance the equations:
2 * (NH4Cl -> NH3 + HCl) = 2 * (50 kJ/mol) = 100 kJ/mol
Step 6: Add the enthalpy changes (ΔH values) for the balanced equations to obtain the enthalpy change for the desired equation:
ΔH3 = ΔH1 + ΔH2' = -59 kJ/mol + 100 kJ/mol = 41 kJ/mol
So, the expected enthalpy change for the reaction NaOH + NH4Cl -> NaCl + NH3 + H2O is 41 kJ/mol.
Based on your calculation of -4.3 kJ/mol, it seems like you forgot to flip equation 2 and multiply it by the required factor to balance the equations. By including these steps, you should have obtained the correct value of 41 kJ/mol.
To calculate the enthalpy change for the reaction NaOH + NH4Cl → NaCl + NH3 + H2O using Hess's Law, we need to apply the concept of enthalpy conservation.
The given mean enthalpy changes for the reactions involved are as follows:
ΔH(NaOH & HCl) = -59 kJ/mol
ΔH(NH3 & HCl) = -50 kJ/mol
ΔH(NaOH & NH4Cl) = -5.7 kJ/mol
Hess's Law states that the enthalpy change for a reaction is the sum of the enthalpy changes for the individual reactions that form the overall reaction. In this case, we can use the given reactions to construct the desired overall reaction by manipulating and combining the reactions' equations.
First, let's manipulate the given reactions to match the overall reaction:
1. Multiply the second equation (NH3 & HCl) by 2:
2NH3 + 2HCl → 2NH3 + 2H2O (ΔH' = -2 * 50 kJ/mol = -100 kJ/mol)
2. Multiply the third equation (NaOH & NH4Cl) by 2:
2NaOH + 2NH4Cl → 2NaCl + 2NH3 + 2H2O (ΔH" = -2 * 5.7 kJ/mol = -11.4 kJ/mol)
Now, let's add the manipulated reactions to get the overall reaction:
2NaOH + 2NH4Cl + 2NH3 + 2HCl → 2NaCl + 2NH3 + 2H2O + 2NH3 + 2H2O
Cancelling out the common species:
2NaOH + 2NH4Cl + 2HCl → 2NaCl + 4NH3 + 4H2O
The overall equation simplifies to:
NaOH + NH4Cl + HCl → NaCl + 2NH3 + 2H2O
Now, let's calculate the enthalpy change for the overall reaction using the manipulated reactions' enthalpy changes:
ΔH overall = ΔH(NaOH & NH4Cl) + ΔH(NH3 & HCl) + (2 * ΔH(NH3 & H2O))
= -11.4 kJ/mol + (-100 kJ/mol) + (2 * 0 kJ/mol) (since ΔH(NH3 & H2O) was not given)
So, ΔH overall = -11.4 kJ/mol - 100 kJ/mol = -111.4 kJ/mol
Based on the calculations, the enthalpy change to be expected for the reaction NaOH + NH4Cl → NaCl + NH3 + H2O is -111.4 kJ/mol. Therefore, the given value of -4.3 kJ/mol is incorrect.